On Wed, Jun 10, 2015 at 04:31:55PM +0200, Petr Mladek wrote:
> > +static void __printk_nmi_flush(struct irq_work *work)
> > +{
> > + struct nmi_seq_buf *s = container_of(work, struct nmi_seq_buf, work);
> > + int len, last_i = 0, i = 0;
> > +
> > +again:
> > + len = seq_buf_used(&s->seq);
> > + if (!len)
> > + return;
> > +
> > + /* Print line by line. */
> > + for (; i < len; i++) {
> > + if (s->buffer[i] == '\n') {
> > + print_seq_line(s, last_i, i);
> > + last_i = i + 1;
> > + }
> > + }
> > + /* Check if there was a partial line. */
> > + if (last_i < len) {
> > + print_seq_line(s, last_i, len - 1);
> > + pr_cont("\n");
> > + }
> > +
> > + /*
> > + * Another NMI could have come in while we were printing
> > + * check if nothing has been added to the buffer.
> > + */
> > + if (cmpxchg_local(&s->seq.len, len, 0) != len)
> > + goto again;
>
> If another NMI comes at this point, it will start filling the buffer
> from the beginning. If it is fast enough, it might override the text
> that we print above.
How so? If the cmpxchg succeeded and len == 0, we flushed everything and
are done with it, if another NMI comes in and 'overwrites' it, that's
fine, right?
> > +static int vprintk_nmi(const char *fmt, va_list args)
> > +{
> > + struct nmi_seq_buf *s = this_cpu_ptr(&nmi_print_seq);
> > + unsigned int len = seq_buf_used(&s->seq);
> > +
> > + irq_work_queue(&s->work);
> > + seq_buf_vprintf(&s->seq, fmt, args);
>
> This is more critical. seq_buf_vprintf() has the following code:
>
> len = vsnprintf(s->buffer + s->len, s->size - s->len, fmt, args);
>
> The race might look like:
>
> CPU0 CPU1
>
No, everything is strictly per cpu.
> To be honest, I am not familiar with cmpxchg_local(). But I think that
> it can't do much difference. The value has to be synced to the other
> CPUs one day.
Nope..
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